1. Reversible reactions and closed systems
A reversible reaction can go in both directions. We show this with the ⇌ symbol. The forward reaction turns reactants (left) into products (right); the reverse reaction turns products back into reactants.
Equilibrium is only reached in a closed system (matter cannot enter or leave; energy may be transferred) under steady conditions (constant temperature, and constant pressure or volume for gases). Kinetics tells you how fast a reaction goes; equilibrium tells you how far it goes.
Not every reversible reaction reaches equilibrium in practice. In an open container a gaseous product can escape, and if the conditions keep changing the mixture never settles. For example, heating calcium carbonate in an open crucible lets CO2 escape, so the decomposition keeps going instead of reaching equilibrium.
AQA 3.1.6 deals with homogeneous equilibria: every species is in the same phase (all gases, or all in one solution or one liquid mixture).
2. Dynamic equilibrium
At dynamic equilibrium:
- the forward and reverse reactions are both still happening, at equal rates;
- so the concentrations of reactants and products stay constant.
Constant does not mean equal. At equilibrium there may be far more products than reactants, or the other way round. The reaction has not stopped: molecules keep reacting in both directions, but each change is cancelled by the opposite one.
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Concentration–time graph approaching equilibrium
Labels to include:
- x-axis: Time; y-axis: Concentration / mol dm⁻³
- Reactant curve starting high and falling, curving to a horizontal line
- Product curve starting at zero and rising, curving to a horizontal line
- The two final horizontal lines at DIFFERENT heights (not meeting)
- Vertical dashed line marking the time equilibrium is reached
- Label: 'equilibrium — concentrations constant, not necessarily equal'
Starting with reactants only, the reactant concentration falls and the product concentration rises. Both curves flatten at the same moment, when equilibrium is reached. The plateaux are at different heights: the curves do not have to cross or meet.
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Rate–time graph approaching equilibrium
Labels to include:
- x-axis: Time; y-axis: Rate
- Forward rate starting high and decreasing
- Reverse rate starting at zero (if no products present at the start) and increasing
- Both lines meeting and continuing as ONE horizontal line above zero
- Label: 'forward rate = reverse rate ≠ 0'
At the start there is no product, so the reverse rate is zero. As reactants are used up the forward rate falls; as products build up the reverse rate rises. At equilibrium the two rates are equal and not zero, which is what makes the equilibrium dynamic. If some product is present at the start, the reverse rate starts above zero.
3. Position of equilibrium vs rate
The position of equilibrium describes the composition of the equilibrium mixture. If the position lies “to the right”, the equilibrium mixture contains mostly products; “to the left” means mostly reactants.
Position and rate are separate ideas. A mixture can reach equilibrium quickly but contain very little product, or slowly but contain a lot. Changing conditions can affect the rate at which equilibrium is reached, the position of equilibrium, or both.
4. Le Chatelier's principle
Le Chatelier's principle: if a change is made to the conditions of a system at equilibrium, the position of equilibrium shifts in the direction that opposes the change.
Two cautions:
- The shift only partly opposes the change. It does not fully restore the original conditions: after heating, the new equilibrium is still hotter than before.
- The system does not “want” or “try” anything. The shift happens because the change affects the forward and reverse rates differently until they are equal again.
5. Changing temperature
- Increasing temperature favours the endothermic direction (it takes in heat, opposing the rise).
- Decreasing temperature favours the exothermic direction (it releases heat, opposing the fall).
The ΔH value quoted with an equation is for the forward direction. If ΔH is negative, the forward reaction is exothermic and the reverse is endothermic.
Temperature is the only one of these changes that also changes the value of Kc. That is covered in Part 3.
6. Changing pressure
Pressure only matters for equilibria involving gases, and only through the number of moles of gas on each side. Count gas moles using the balancing numbers.
- Increasing the pressure by compressing the mixture at constant temperature favours the side with fewer moles of gas.
- Decreasing pressure favours the side with more moles of gas.
- If both sides have the same number of gas moles, changing pressure does not change the composition of the equilibrium mixture (though compression still raises every concentration and the rates).
| Equation | Gas moles left | Gas moles right | Higher pressure favours |
|---|---|---|---|
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | 4 | 2 | Right (forward) |
| N₂O₄(g) ⇌ 2NO₂(g) | 1 | 2 | Left (reverse) |
| H₂(g) + I₂(g) ⇌ 2HI(g) | 2 | 2 | No change in composition |
7. Adding an inert gas
Adding an unreactive gas such as argon to a fixed volume at constant temperature raises the total pressure. But the amount of each reacting gas per unit volume does not change, so their concentrations (and partial pressures) are unchanged. There is no shift in the position of equilibrium.
If instead the inert gas is added while the total pressure is kept fixed, the container must expand. The reacting gases are diluted, which favours the side with more gas moles, just like lowering the pressure. This is an extension of the concentration reasoning; Kp calculations belong to a later topic.
8. Changing concentration
With all other conditions kept fixed:
- adding a reactant, or removing a product, favours the forward reaction;
- adding a product, or removing a reactant, favours the reverse reaction.
Diluting a solution changes every concentration at once. Only discuss dilution when the question gives the volume change explicitly, and treat it as its own case.
9. Catalysts
A catalyst increases the rates of the forward and reverse reactions by the same factor. So equilibrium is reached sooner, but the equilibrium composition is the same. A catalyst does not shift the position of equilibrium and does not change Kc.
So do not write that a catalyst “raises the yield”. At the same temperature and pressure, the equilibrium yield is unchanged. (In industry a catalyst can allow a lower temperature to be used; that is covered in Part 2.)
10. Worked example: NO₂ and N₂O₄
N2O4 is colourless and NO2 is brown. The forward reaction is endothermic.
Cooling
Lowering the temperature favours the exothermic direction, which is the reverse here. The position shifts to the left, less NO2 is present, so the mixture becomes paler.
Warming
Raising the temperature favours the endothermic forward direction. The position shifts right, more NO2 forms, and the mixture becomes browner.
Compressing a sealed syringe (constant temperature)
Observe through the same optical path length (for example, across the syringe barrel). Two separate things happen:
- Immediately: the same gas is squeezed into a smaller volume. The concentration of NO2 rises at once, so the colour looks darker. This is a density effect; the composition has not changed yet.
- Then: higher pressure favours the side with fewer gas moles (1 mol N2O4 vs 2 mol NO2). The position shifts left, some NO2 is converted to N2O4, and the colour partly fades.
When viewed through the same optical path length, the final colour is darker than before compression because [NO₂] remains higher. Changing the viewing path can also affect the apparent colour.
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Sealed syringe of NO₂/N₂O₄ before and after compression
Labels to include:
- Three syringe panels side by side
- Panel 1: 'Equilibrium at original volume' — medium brown
- Panel 2: 'Immediately after compression' — plunger pushed in, darker brown (same composition, higher concentration)
- Panel 3: 'New equilibrium' — slightly paler than panel 2 but darker than panel 1
- Arrow between panels 2 and 3: 'position shifts left, towards fewer gas moles'
- Note: constant temperature throughout
The darkening in panel 2 comes only from squeezing the same gas into a smaller volume. The partial fading in panel 3 is the equilibrium shift towards N2O4. Panel 3 is still darker than panel 1.
Quick checks
These are Finesse practice questions. The step-by-step answers are indicative worked solutions, not official AQA mark allocations.
Q1. State two features of a reaction at dynamic equilibrium, and explain why “the concentrations of reactants and products are equal” is wrong.Show answer
Feature 1: the forward and reverse reactions happen at equal rates (both still occurring).
Feature 2: the concentrations of reactants and products stay constant.
The concentrations are constant but usually not equal; the equilibrium mixture can be mostly reactants or mostly products.
Q2. For 2NO2(g) ⇌ N2O4(g), ΔH = −58 kJ mol−1. A sealed flask of the mixture is placed in hot water. Explain what happens to the colour. (NO2 brown, N2O4 colourless.)Show answer
Change: temperature increases.
Favoured direction: endothermic. The forward reaction is exothermic (ΔH negative), so the reverse direction is endothermic.
Shift: the position moves to the left, so more NO2 forms and the mixture becomes browner.
Q3. A mixture of H2(g), I2(g) and HI(g) at equilibrium is compressed to half its volume at constant temperature. Does the position of equilibrium shift? Explain what happens to the purple colour of I2.Show answer
Gas moles: 2 on the left (1 H2 + 1 I2), 2 on the right (2 HI). They are equal, so the position of equilibrium does not shift and the composition is unchanged.
The concentration of I2 doubles because the same amount is in half the volume, so the purple colour looks darker. This is not a shift.
Q4. Argon is added to a fixed-volume vessel containing N2, H2 and NH3 at equilibrium, at constant temperature. The total pressure rises. A student says the yield of NH3 increases because pressure has increased. Evaluate this.Show answer
The student is wrong. The volume is fixed, so the amounts of N2, H2 and NH3 per unit volume (their concentrations and partial pressures) are unchanged.
The forward and reverse rates are therefore unchanged, so there is no shift and the yield of NH3 stays the same.
Compression changes the reacting gases’ concentrations and can shift this equilibrium; a temperature change can also shift it.
Q5. For Fe3+(aq) + SCN−(aq) ⇌ [FeSCN]2+(aq) (red), a reagent is added that removes Fe3+ ions from solution, at constant temperature with negligible volume change. Predict and explain the colour change.Show answer
Change: the concentration of the reactant Fe3+ decreases.
Favoured direction: the direction that makes more Fe3+, which is the reverse reaction.
Shift: the position moves to the left, so [FeSCN]2+ decreases and the red colour becomes paler.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.6 Equilibria revision guide (N. Goalby, Oct 2025 upload) — Primary checklist for dynamic equilibrium, Le Chatelier's principle and catalysts. We qualify its claims that all reversible reactions reach equilibrium and that the system 'tries to reverse' a change.
- AQA 7405 specification — 3.1.6 Chemical equilibria, Le Chatelier's principle and Kc — 3.1.6.1 Chemical equilibria and Le Chatelier's principle (homogeneous reactions; catalyst does not affect position).
- AQA 7404/1 mark scheme, June 2023 — Q08.3: decreased temperature → shift in exothermic direction → opposes the decrease.
- AQA 7404/1 examiner report, June 2023 — Q08.3: refer explicitly to a shift in the position of equilibrium.
- AQA 7404/1 mark scheme, June 2022 — Q07.3: catalyst increases forward and reverse rates equally.
- AQA 7404/1 examiner report, June 2022 — Q07.3: 'increases the rate' or 'no effect on position' alone did not score.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.